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erik [133]
3 years ago
12

Predict the major products when isopropylbenzene is irradiated by light and treated with excess br2.

Chemistry
1 answer:
WARRIOR [948]3 years ago
5 0
The reaction of isopropylbenzene with bromine under light is a radical reaction. The light initially reactions with the bromine in the initiation step to form the bromine radicals:

Br₂ + hv → 2Br·

The next step is the first propagation step where a bromine radical reacts with the isopropylbenzene to abstract the hydrogen from the tertiary carbon in the isopropyl group:

                CH₃                          CH₃
                 |                               |
Br· + Ph--C--H  → H-Br + Ph--C·
                 |                               |
                CH₃                          CH₃

We have now formed a radical in the benzylic position of the isopropylbenzene structure. Now this radical will react with another molecule of bromine in a second propogation step to form the final brominated product:

       CH₃                        CH₃
        |                             |
Ph--C·   + Br-Br → Ph--C--Br  + Br·
        |                             |
       CH₃                        CH₃

The bromine radical can terminate in a reaction with another bromine radical or the benzylic radical to give more of the final product. The final product is shown and is called (2-bromopropan-2-yl)benzene.
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4 0
3 years ago
3.65 gram of hcl is dissolved in 180 gram of water. Find the total number of molecules of hydrogen​
Morgarella [4.7K]

Answer:

Molec_{\ H_{tot}}=1.206x10^{25}molec

Explanation:

Hello.

In this case, taking into account that HCl has one molecule of hydrogen per mole of compound which weights 36.45 g/mol, we compute the number of molecules of hydrogen in hydrochloric acid by considering the given mass and the Avogadro's number:

molec_{\ H}=3.65gHCl*\frac{1molHCl}{36.45gHCl} *\frac{1molH}{1molHCl}*\frac{6.022x10^{23}molec_\ H}{1molH}  =6.03x10^{22}molec

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molec_{\ H}=180gH_2O*\frac{1molH_2O}{18gH_2O} *\frac{2molH}{1molH_2O}*\frac{6.022x10^{23}molec_\ H}{1molH}  =1.20x10^{25}molec

Thus, the total number of molecules turns out:

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Regards.

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